Mathematics for Computer Science Exercises from Chapter 3

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1 Mathematics for Computer Science Exercises from Chapter 3 Silvio Capobianco Last update: 19 September 2018 Problems from Section 3.1 Problem 3.2. Your class has a textbook and a final exam. Let P, Q, and R be the following propositions: P ::= You get an A on the final exam. Q ::= You do every exercise in the book. R ::= You get an A in the class. Translate following assertions into propositional formulas using P, Q, R, and the propositional connectives and, not(), implies. (a) You get an A in the class, but you do not do every exercise in the book. (b) You get an A on the final exam, you do every exercise in the book, and you get an A in the class. (c) To get an A in the class, it is necessary for you to get an A on the final. (d) You get an A on the final, but you don t do every exercise in this book; nevertheless, you get an A in this class. 1

2 Solution. (a) In this case, R is verified, Q is not, and P is irrelevant: the assertion translates as R and not(q). (b) Here P, Q, and R are all verified, so this assertion translates as P and Q and R. Recall that and is associative, so (P and Q) and R is equivalent to P and (Q and R). (c) Here we have a clear implication, and a causal one too! What the assertion says, is that if you get an A in the class, it means that you had gotten an A in the final: the translation in mathematical language is then R implies P, or equivalently, but more menacingly, P or not(r). (d) In this case, P is true, Q is false, and R is true: the assertion translates as P and not(q) and R. At the end of the exercise, observe how, in mathematical language, but and nevertheless mean the same as and. The differences in tone of the three words are lost in translation. Problem 3.5. Sloppy Sam is trying to prove a certain proposition P. He defines two related propositions Q and R, and then proceeds to prove three implications: He then reasons as follows: P implies Q, Q implies R, R implies P. If Q is true, then since I proved Q implies R, I can conclude that R is true. Now, since I proved R implies P, I can conclude that P is true. Similarly, if R is true, then P is true and so Q is true. Likewise, if P is true, then so are Q and R. So any way you look at it, all three of P, Q and R are true. (a) Exhibit truth tables for and for (P implies Q) and (Q implies R) and (R implies P ) (1) P and Q and R. (2) Use these tables to find a truth assignment for P, Q, R so that (1) is T and (2) is F. 2

3 (b) You show these truth tables to Sloppy Sam and he says OK, I m wrong that P, Q and R all have to be true, but I still don t see the mistake in my reasoning. Can you help me understand my mistake? How would you explain to Sammy where the flaw lies in his reasoning? Solution. (a) We first construct the truth table for P and Q and R, as it is almost immediate: For the formula P Q R P and Q and R T T T T T T F F T F T F T F F F F T T F F T F F F F T F F F F F S = (P implies Q) and (Q implies R) and (R implies P ) we proceed in two steps: first, we construct the truth values for each of the implications; then, we compute those for their conjunction. P Q R P implies Q Q implies R R implies P S T T T T T T T T T F T F T F T F T F T T F T F F F T T F F T T T T F F F T F T F T F F F T T T F F F F F T T T T We then see that, if P, Q and R are all F, then (1) is T and (2) is F. (b) Sam is tacitly assuming that some of P, Q and R are true. But why should it be so? His application of the proof-by-cases method is sloppy: the case of P, Q, and R all false was not covered by the others, and should have been considered separately. All Sam has proved is that they are equivalent: 3

4 either they all all true, or all false. To check which is the case, he must find a proof or disproof of any of the three, but at least one of them, which does not depend on the others, but only on other things which he knows, not just assumes, to be true. Problems for Section 3.3 Problem Indicate whether each of the following propositional formulas is valid (V), satisfiable but not valid (S), or not satisfiable (N). For the satisfiable ones, indicate a satisfying truth assignment. M implies Q M implies (P or Q) M implies (M and (P implies M)) (P or Q) implies Q (P or Q) implies (P and Q) (P or Q) implies (M and (P implies M)) (P xor Q) implies Q (P xor Q) implies (P or Q) (P xor Q) implies (M and (P implies M)) Solution. Let s check one by one: S. M implies Q is T as soon as M is F, but is F if M is T and Q is F. S. M implies (P or Q) is T as soon as M is F, but is F if M, P and Q are all T. V. M implies (M and (P implies M)) is valid. If M is F, then M implies (M and (P implies M)) is T; if M is T, then so is P implies M, and so is M and (P implies M). S. (P or Q) implies Q is T if Q is T, but F if P is T and Q is F. S. (P or Q) implies (P or Q) is T if P and Q are both F, but F if they are both T. 4

5 S. (P or Q) implies (M and (P implies M)) is T if P and Q are both F, but F if they are both T and M is F. S. (P xor Q) implies Q is T if P and Q are both T, but F if P is T and Q is F. V. (P or Q) implies (P or Q) is valid. If P and Q are both T or both F, then the implication from F is T; if one is T and the other is F, then one of P and Q is T, and the implication to T is T. S. (P xor Q) implies (M and (P implies M)) is T if P and Q are both F, but F if P is T and Q and M are both F. Problem This problem examines whether the following specifications are satisfiable: 1. If the file system is not locked, then (a) new messages will be queued. (b) new messages will be sent to the messages buffer. (c) the system is functioning normally, and conversely, if the system is functioning normally, then the file system is not locked. 2. If new messages are not queued, then they will be sent to the messages buffer. 3. New messages will not be sent to the message buffer. (a) Begin by translating the five specifications into propositional formulas using four propositional variables: L ::= file system locked, Q ::= new messages are queued, B ::= new messages are sent to the message buffer, N ::= system functioning normally. 5

6 (b) Demonstrate that this set of specifications is satisfiable by describing a single truth assignment for the variables L, Q, B, N and verifying that under this assignment, all the specifications are true. (c) Argue that the assignment determined in part () is the only one that does the job. Solution. (a) Let us rewrite the three specifications as three Boolean formulas α, β and γ: 1. α ::= (not(l) implies (Q and B and N)) and (N implies not(l)). 2. β ::= not(q) implies B. 3. γ ::= not(b). (b) We must find a truth assignment to L, Q, B and N that makes each of α, β and γ take value T. We immediately observe that γ = T if and only if B = F. In this case, for β to be T it must be not(q) = F, hence Q = T. Now, to have α = T, we must have both and not(l) implies (Q and B and N) = T N implies not(l) = T. But the conjunction in the right-hand side of the implication in the first formula is F, because so is B: we must then have not(l) = F, that is, L = T. Then the second formula can only be T if N = F. We then have that the specification is verified if, and only if: 1. the system is locked, 2. new messages are queued, 3. new messages are not sent to the message buffer, and 4. the system does not function normally. (c) Think of our proof of satisfiability as a game where we win if we reach an assignment of truth values which makes all the specifications true, and lose if we make a truth value assignment which makes some specification 6

7 false. Then our first move was forced; our second move was forced, given our first move; and our third move was forced, given the previous two. We had no other way to win the game, so the assignment we obtained was the only winning position, and the only way to satisfy all the specifications. Note that the text of the exercise as reported in the book is imprecise. With that formulation, α becomes: not(l) implies (Q and B and N and (N implies not(l))) But this formula is not equivalent to (not(l) implies (Q and B and N)) and (N implies not(l)), because the former is satisfied with L = Q = T and B = F regardless of the value of N. Problems for Section 3.4 Problem A half dozen different operators may appear in propositional formulas, but just and, or, and not are enough to do the job. That is because each of the operators is equivalent to a simple formula using only these three operators. For example, A implies B is equivalent to not(a) or B. So all occurences of implies in a formula can be replaced using just not and or. (a) Write formulas using only and, or, not() that are equivalent to each of A iff B and xor B. Conclude that every propositional formula is equivalent to an and - or -not formula. (b) Explain why you don t even need and. (c) Explain how to get by with the single operator nand where A nand B is equivalent by definition to not(a and B). Solution. (a) A iff B is equivalent to: while A xor B is equivalent to: (A and B) or (not(a) and not(b)), (A or B) and (not(a) or not(b)). 7

8 Observe how we arrived from the formula for iff to the one for xor by swapping the roles of and and or : this is not a case, as A xor B is equivalent to not(a iff B). (b) That we don t even need and follows from de Morgan s law: not(a and B) not(a) or not(b), which, together with the double negation rule, gives: A and B not(not(a) or not(b)). (c) Because of the previous points, it is sufficient to show that not() and or can be reconstructed by only using nand. One part is easy: not(a) A nand A. For the second one, we use de Morgan s law together with double negation: A or B not(not(a) and not(b)) not(a) nand not(b) (A nand A) nand (B nand B). Problem 3.19(a). Let P be the proposition depending on the propositional variables A, B, C, D whose truth table is: A B C D P T T T T T T T T F F T T F T T T T F F F T F T T T T F T F T T F F T T T F F F T A B C D P F T T T T F T T F F F T F T T F T F F F F F T T F F F T F F F F F T T F F F F T Write out a disjunctive normal form for P. 8

9 Solution. We isolate the rows of the truth table where P is true: A B C D P T T T T T T T F T T T F T T T T F T F T T F F T T T F F F T F T T T T F T F T T F F F T T F F F F T Then, we convert each T/F sequence into a conjunction of variables and their negations, according to the rule seen in class: A B C D conjunction T T T T A B C D T T F T A B C D T F T T A B C D T F T F A B C D T F F T A B C D T F F F A B C D F T T T A B C D F T F T A B C D F F F T A B C D F F F F A B C D Finally, we construct the disjunctive normal form as a disjunction of all those 9

10 conjunctions: P (A B C D) Problems for Section 3.6 Problem For each of the following propositions: 1. x. y. 2x y = 0 2. x. y. x 2y = 0 3. x. (x < 10 implies ( y. (y < x implies y < 9))) 4. x. y. (y > x z. y + z = 100) determine which propositions are true when the variables range over: (a) the nonnegative integers, (b) the integers, (c) the real numbers. Solution. (a) If the variables are nonnegative integers: 1. is true, because however given x, we can choose y = 2x. 10

11 2. is false, because for x = 1 the difference 1 2y is odd for every integer y, and cannot be zero. 3. is true, because if m and n are any two integers, then m < n if and only if m n is false, because if we choose x = 101, y > x, and z any nonnegative integer, then y + z > 100. (b) If the variables are integers: 1. is true, because however given x, we can choose y = 2x. 2. is false, because for x = 1 the difference 1 2y is odd for every integer y, and cannot be zero. 3. is true, because if m and n are any two integers, then m < n if and only if m n is true, because however we choose x and y > x, we can always set z = 100 y, which is integer if y is. (c) If the variables are real numbers: 1. is true, because however given x, we can choose y = 2x. 2. is true, because however given x, we can choose y = x/2. 3. is false, because for x = 9.75 we can take y = 9.5, which is smaller than x but larger than is true, because however we choose x and y > x, we can always set z = 100 y. Countermodels Definition. A countermodel for a predicate formula P is an environment for its variables, together with an interpretation of P within that environment, which makes P false. 11

12 For example, the formula: x. y. P (x, y) has a countermodel where the environment is the set of the natural numbers, and P (x, y) ::= x > y. Then the formula becomes: x N. y N. x > y, which is false, because for x = 0 there is no y N such that 0 > y. Problem Find a counter-model showing the following is not valid. x. P (x) implies x. P (x) (Just define your counter-model. You do not need to verify that it is correct.) Solution. Recall that quantifiers bind stronger than propositional connectives: therefore, the formula means ( x. P (x)) implies ( x. P (x)). Consider a situation where: the variables range over the nonnegative integers, and P (x) ::= x = 17. Then indeed there exists a nonnegative integer which is 17, but there is also a nonnegative integer which is not 17. Problem Find a counter-model showing the following is not valid. ( x. P (x) and x. Q(x)) implies x. (P (x) and Q(x)) (Just define your counter-model. You do not need to verify that it is correct.) Solution. We can recycle the solution of Problem 3.25: the variables range over the nonnegative integers, 12

13 P (x) ::= x = 17, and Q(x) ::= x = 16. Then the antecedent of the implication is true, because we can substitute x = 17 in the left-hand side of the and and x = 16 in the right-hand side. But the consequent is false, because no nonnegative integer is equal to both 17 and 16. Problem 3.28 (tweaked). (a) Verify that the propositional formula is valid. (P implies Q) or (Q implies P ) (b) The valid formula of part (a) leads to sound proof method: to prove that an implication is true, just prove that its converse is false. 1 For example, from middle school arithmetics we know that the assertion If an integer is not prime, then it is negative is false (4 is not prime, and positive). This allows us to reach at the correct conclusion that its converse, If an integer is negative, then it is not prime is true, as indeed it is (prime numbers are all larger than 1). But wait a minute! The implication If an integer is prime, then it is negative is completely false. So we could conclude that its converse If an integer is negative, then it is prime 1 This problem was stimulated by the discussion of the fallacy in J. Beam, A Powerful Method of Non-Proof, The College Mathematics Journal 48(1),

14 should be true, but in fact the converse is also completely false. So something has gone wrong here. Explain what. Solution. (a) This is easily done via truth table: P Q (P implies Q) or (Q implies P ) T T T T T T F F T T F T T T F F F T T T Alternatively: if P implies Q is true, then the formula is true; if P implies Q is false, then P is true and Q is false, which makes Q implies P true, and the formula true as well. (b) The formula which we proved valid is a propositional formula: it is true whenever P and Q have a definite truth value. This allows us to conclude, for example, that x. [(P (x) implies Q(x)) or (Q(x) implies P (x))] (3) is valid: as soon as the value of x is chosen, so are the truth values of P (x) and Q(x), and whatever those are, the proposition in square brackets is true. However, If an integer is prime, then it is negative is not a propositional formula: it is a predicate formula, which is correctly translated into predicate logic as x Z. (isprime(x) implies x < 0). (Z, which comes from the German word Zahlen, meaning numbers, is the standard notation for the set of integer numbers.) Now, when we start from the falsity of to prove If an integer is prime, then it is negative If an integer is negative, then it is prime 14

15 we are not applying (3), but a different formula: x. (P (x) implies Q(x)) or x. (Q(x) implies P (x)) (4) where we are giving to P (x) the meaning x is prime and to Q(x) the meaning x is negative. But the formula (4) is not valid in predicate logic, and our interpretation with being negative and being prime in the context of integer numbers is precisely a counter-model to it! We can also consider the counter-model of Problem 3.26: 1. for x = 17 we have P (17) true and Q(17) false, so x. (P (x) implies Q(x)) is false in this model; 2. for x = 16 we have Q(16) true and P (17) false, so x. (Q(x) implies P (x)) is also false in this model. 15